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Why do fundamental circuit laws break down at high frequency AC?
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We're just starting the whole RF scene having dealt with DC and low frequency AC for all our previous courses.
I understand that at high frequency AC, fundamental circuit laws don't apply anymore and the classic passive component models need to be changed. The justification for this was that at high frequency AC transmission, the wavelength becomes much smaller and can sometimes be smaller than the wiring on PCBs etc.
I understand that this is an issue when transmitting through free space with electromagnetic waves but why is this an issue with actual physical wires and PCBs being driven by an AC source? I mean it's a direct connection, we aren't using electromagnetic waves to propogate through free space and so wavelength and stuff shouldn't matter right?
ac high-frequency
asked 1 hour ago
AlfroJang80
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favorite
We're just starting the whole RF scene having dealt with DC and low frequency AC for all our previous courses.
I understand that at high frequency AC, fundamental circuit laws don't apply anymore and the classic passive component models need to be changed. The justification for this was that at high frequency AC transmission, the wavelength becomes much smaller and can sometimes be smaller than the wiring on PCBs etc.
I understand that this is an issue when transmitting through free space with electromagnetic waves but why is this an issue with actual physical wires and PCBs being driven by an AC source? I mean it's a direct connection, we aren't using electromagnetic waves to propogate through free space and so wavelength and stuff shouldn't matter right?
ac high-frequency
asked 1 hour ago
AlfroJang80
435
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We're just starting the whole RF scene having dealt with DC and low frequency AC for all our previous courses.
I understand that at high frequency AC, fundamental circuit laws don't apply anymore and the classic passive component models need to be changed. The justification for this was that at high frequency AC transmission, the wavelength becomes much smaller and can sometimes be smaller than the wiring on PCBs etc.
I understand that this is an issue when transmitting through free space with electromagnetic waves but why is this an issue with actual physical wires and PCBs being driven by an AC source? I mean it's a direct connection, we aren't using electromagnetic waves to propogate through free space and so wavelength and stuff shouldn't matter right?
ac high-frequency
asked 1 hour ago
AlfroJang80
435
We're just starting the whole RF scene having dealt with DC and low frequency AC for all our previous courses.
I understand that at high frequency AC, fundamental circuit laws don't apply anymore and the classic passive component models need to be changed. The justification for this was that at high frequency AC transmission, the wavelength becomes much smaller and can sometimes be smaller than the wiring on PCBs etc.
I understand that this is an issue when transmitting through free space with electromagnetic waves but why is this an issue with actual physical wires and PCBs being driven by an AC source? I mean it's a direct connection, we aren't using electromagnetic waves to propogate through free space and so wavelength and stuff shouldn't matter right?
ac high-frequency
ac high-frequency
asked 1 hour ago
AlfroJang80
435
asked 1 hour ago
AlfroJang80
435
asked 1 hour ago
AlfroJang80
435
asked 1 hour ago
AlfroJang80
435
asked 1 hour ago
AlfroJang80
435
435
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add a comment |Â
4 Answers
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up vote
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Actually, it is all about the waves. Even when dealing with DC, it is all managed by the electrical and magnetic fields and waves.
The "fundamental laws" aren't breaking down. The rules you have learned are simplifications that deliver accurate answers under certain conditions - you haven't yet learned the fundamental laws. You are about to learn the fundamental laws after having used simplifcations.
Part of the assumed conditions for the simplified rules is that the circuit is much smaller than the wave length of signal(s) involved. In those conditions, you can assume that a signal is in the same state across the circuit. That leads to a lot of simplifications in the equations describing the circuit.
As the frequencies get higher (or the circuits larger) so that the circuit is an appreciable fraction of the wavelength, that assumption is no longer valid.
The effects of wavelength on the operation of electrical circuits first became obvious at low frequencies but with very large circuits - telegraph lines.
When you start working with RF, you reach wavelengths such that the size of a circuit that sits on your desk is an appreciable fraction of the wavelength of the signals used.
So, you start having to pay attention to things you could conveniently ignore before.
The rules and equations you are now learning also apply to simpler, lower frequency circuits. You can use the new things to solve the simpler circuits- you just have to have more information and solve more complicated equations.
answered 51 mins ago
JRE
17.6k43260
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up vote
1
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Electrical signals take time to propagate through wires (and PCB traces). Slower than EM waves through a vacuum or air, always.
For example a twisted pair in a CAT5e cable has a velocity factor of 64%, so the signal travels at 0.64c, and it will go about 8" in a nanosecond. A nanosecond is a long time in some electronic contexts.It's 4 clock cycles in a modern CPU, for example.
Any configuration of conductors of finite size has inductance and capacitance and (usually) resistance so it can be approximated using lumped components at a finer level of granularity. You might replace the wire with 20 series inductors and resistors with 20 capacitors to the ground plane. If the wavelength is very short compared to the length, you might need 200 or 2000 or .. whatever to closely approximate the wire and other methods might start to look attractive, such as transmission line theory (typically a one semester undergrad course for EEs).
"Laws" like KVL, KCL are mathematical models that approximate reality very accurately under appropriate conditions. More general laws such as Maxwell's equations apply more generally. There might be some situations (relativistic perhaps) where Maxwell's equations are no longer very accurate.
answered 1 hour ago
Spehro Pefhany
194k4139384
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up vote
1
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I mean it's a direct connection, we aren't using electromagnetic waves to propogate through free space and so wavelength and stuff shouldn't matter right?
That's a very wrong assumption. The signals are still EM waves and remain EM waves, if they propagate through free space or a conductor. The laws remain the same.
At connections (wires) in the order of the length of the wavelength you can no longer used the "lumped element" approach. The "lumped element" approach means that connections are considered "ideal". For high frequency signals at distances in the order of the wavelength and larger, this approach is invalid.
So remember: the EM laws do not change as an EM wave travels through space or a conductor, they apply in both cases. EM waves remain EM waves in free space or in a conductor.
answered 1 hour ago
Bimpelrekkie
42.4k23791
Okay. I understand that the EM waves still exist when transmitting AC voltages through a wire - but they don't contribute to the actual current flow right (apart from reducing it a bit with the opposing emf). So then why should we abandon all our low-frequency and DC models when essentially the AC current is still flowing fine through that wire. I just don't see how the wavelength being too small comes into play when we have a direct wire from the AC source and load.
– AlfroJang80
1 hour ago
One should add, even for the most-high-speed signals one could expect on a "normal" PCB, the lumped model is still applicable if capacitance and inductance of a whole track are taken into account. The distances are small, after all.
– Janka
1 hour ago
@AlfroJang80: This becomes crucial as soon you are in the UHF range. Tracks acts as antennas then.
– Janka
59 mins ago
1
@AlfroJang80, a dipole antenna is just a pair of direct wires from the feed to their open ends. And yet it can transmit and receive wireless RF signals. Somewhere between a very short wire that doesn't transmit or receive any energy, and a quarter wave dipole that transmits and receives very efficiently, there must be a middle ground where radiation effects are significant but not dominant.
– The Photon
40 mins ago
Okay. Makes much more sense now. I think my confusion is that with an AC source and a wire. Is the electromagnetic field essential to current flowing? Up to now, i think of it as the emag. wave as being a seperate thing that is emitted when AC current flows through a wire. Is that correct? Or is the emag wave the only thing that makes current flow through the wire?
– AlfroJang80
2 mins ago
add a comment |Â
up vote
1
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They don't break down, but when the rise time approaches 10% or is less than the propagation delay to a load impedance matching is important due to that wavelength. Load Impedance is inverted to a source at 1/4 wavelength whether it is conducted or radiated.
If the load is not matched impedance to the "transmission line and source" reflections will occur according to some coefficient called return loss and the reflection coefficient.
Here is an experiment you can do to demonstrate conducted EM waves.
If you try probing a 1MHz square wave on a 10:1 scope probe with the 10cm ground clip you might see 20MHz lumped coax resonance. Yes, the probe is not matched to the 50 Ohm generator so reflections will occur according to the 10nH/cm ground lead and 50pF/m special probe coax. Still a lumped element (LC) response.
Reducing the 10:1 probe to <1cm to just the pin tip and ring without long ground clip, raises the resonant frequency perhaps to the limitation of the probe and scope at 200MHz.
NOw try a 1:1 1m coax which is 20ns/m so a 20~50MHz square wave on a 1m coax with a 1:1 probe will see a reflection at one fraction of a wavelength and horrible square wave response unless terminated at the scope with 50 Ohms. This is a conducted EM wave reflection.
But consider a fast logic signal with 1ns rise time may have 25 Ohm source impedance and it has a >300MHz BW so overshoot can be a measurement error or actual impedance mismatch with track length reflections.
Now compute 5% of the wavelength of 300MHz at 3e8m/s for air and 2e8m/s for coax and see what the propagation delay times are that cause echoes from mismatched load e.g. CMOS high Z and say 100-ohm tracks. This is why controlled impedances are needed usually above 20~50MHz and this as an effect on ringing or overshoot or impedance mismatch. But without, this is why logic has a such a large grey zone between "0 & 1" to allow for some ringing.
If any words are unknown, look them up.
answered 1 hour ago
Tony EE rocketscientist
57.6k22084
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Actually, it is all about the waves. Even when dealing with DC, it is all managed by the electrical and magnetic fields and waves.
The "fundamental laws" aren't breaking down. The rules you have learned are simplifications that deliver accurate answers under certain conditions - you haven't yet learned the fundamental laws. You are about to learn the fundamental laws after having used simplifcations.
Part of the assumed conditions for the simplified rules is that the circuit is much smaller than the wave length of signal(s) involved. In those conditions, you can assume that a signal is in the same state across the circuit. That leads to a lot of simplifications in the equations describing the circuit.
As the frequencies get higher (or the circuits larger) so that the circuit is an appreciable fraction of the wavelength, that assumption is no longer valid.
The effects of wavelength on the operation of electrical circuits first became obvious at low frequencies but with very large circuits - telegraph lines.
When you start working with RF, you reach wavelengths such that the size of a circuit that sits on your desk is an appreciable fraction of the wavelength of the signals used.
So, you start having to pay attention to things you could conveniently ignore before.
The rules and equations you are now learning also apply to simpler, lower frequency circuits. You can use the new things to solve the simpler circuits- you just have to have more information and solve more complicated equations.
answered 51 mins ago
JRE
17.6k43260
add a comment |Â
up vote
3
down vote
Actually, it is all about the waves. Even when dealing with DC, it is all managed by the electrical and magnetic fields and waves.
The "fundamental laws" aren't breaking down. The rules you have learned are simplifications that deliver accurate answers under certain conditions - you haven't yet learned the fundamental laws. You are about to learn the fundamental laws after having used simplifcations.
Part of the assumed conditions for the simplified rules is that the circuit is much smaller than the wave length of signal(s) involved. In those conditions, you can assume that a signal is in the same state across the circuit. That leads to a lot of simplifications in the equations describing the circuit.
As the frequencies get higher (or the circuits larger) so that the circuit is an appreciable fraction of the wavelength, that assumption is no longer valid.
The effects of wavelength on the operation of electrical circuits first became obvious at low frequencies but with very large circuits - telegraph lines.
When you start working with RF, you reach wavelengths such that the size of a circuit that sits on your desk is an appreciable fraction of the wavelength of the signals used.
So, you start having to pay attention to things you could conveniently ignore before.
The rules and equations you are now learning also apply to simpler, lower frequency circuits. You can use the new things to solve the simpler circuits- you just have to have more information and solve more complicated equations.
answered 51 mins ago
JRE
17.6k43260
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Actually, it is all about the waves. Even when dealing with DC, it is all managed by the electrical and magnetic fields and waves.
The "fundamental laws" aren't breaking down. The rules you have learned are simplifications that deliver accurate answers under certain conditions - you haven't yet learned the fundamental laws. You are about to learn the fundamental laws after having used simplifcations.
Part of the assumed conditions for the simplified rules is that the circuit is much smaller than the wave length of signal(s) involved. In those conditions, you can assume that a signal is in the same state across the circuit. That leads to a lot of simplifications in the equations describing the circuit.
As the frequencies get higher (or the circuits larger) so that the circuit is an appreciable fraction of the wavelength, that assumption is no longer valid.
The effects of wavelength on the operation of electrical circuits first became obvious at low frequencies but with very large circuits - telegraph lines.
When you start working with RF, you reach wavelengths such that the size of a circuit that sits on your desk is an appreciable fraction of the wavelength of the signals used.
So, you start having to pay attention to things you could conveniently ignore before.
The rules and equations you are now learning also apply to simpler, lower frequency circuits. You can use the new things to solve the simpler circuits- you just have to have more information and solve more complicated equations.
answered 51 mins ago
JRE
17.6k43260
Actually, it is all about the waves. Even when dealing with DC, it is all managed by the electrical and magnetic fields and waves.
The "fundamental laws" aren't breaking down. The rules you have learned are simplifications that deliver accurate answers under certain conditions - you haven't yet learned the fundamental laws. You are about to learn the fundamental laws after having used simplifcations.
Part of the assumed conditions for the simplified rules is that the circuit is much smaller than the wave length of signal(s) involved. In those conditions, you can assume that a signal is in the same state across the circuit. That leads to a lot of simplifications in the equations describing the circuit.
As the frequencies get higher (or the circuits larger) so that the circuit is an appreciable fraction of the wavelength, that assumption is no longer valid.
The effects of wavelength on the operation of electrical circuits first became obvious at low frequencies but with very large circuits - telegraph lines.
When you start working with RF, you reach wavelengths such that the size of a circuit that sits on your desk is an appreciable fraction of the wavelength of the signals used.
So, you start having to pay attention to things you could conveniently ignore before.
The rules and equations you are now learning also apply to simpler, lower frequency circuits. You can use the new things to solve the simpler circuits- you just have to have more information and solve more complicated equations.
answered 51 mins ago
JRE
17.6k43260
answered 51 mins ago
JRE
17.6k43260
answered 51 mins ago
JRE
17.6k43260
answered 51 mins ago
JRE
17.6k43260
17.6k43260
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add a comment |Â
up vote
1
down vote
Electrical signals take time to propagate through wires (and PCB traces). Slower than EM waves through a vacuum or air, always.
For example a twisted pair in a CAT5e cable has a velocity factor of 64%, so the signal travels at 0.64c, and it will go about 8" in a nanosecond. A nanosecond is a long time in some electronic contexts.It's 4 clock cycles in a modern CPU, for example.
Any configuration of conductors of finite size has inductance and capacitance and (usually) resistance so it can be approximated using lumped components at a finer level of granularity. You might replace the wire with 20 series inductors and resistors with 20 capacitors to the ground plane. If the wavelength is very short compared to the length, you might need 200 or 2000 or .. whatever to closely approximate the wire and other methods might start to look attractive, such as transmission line theory (typically a one semester undergrad course for EEs).
"Laws" like KVL, KCL are mathematical models that approximate reality very accurately under appropriate conditions. More general laws such as Maxwell's equations apply more generally. There might be some situations (relativistic perhaps) where Maxwell's equations are no longer very accurate.
answered 1 hour ago
Spehro Pefhany
194k4139384
add a comment |Â
up vote
1
down vote
Electrical signals take time to propagate through wires (and PCB traces). Slower than EM waves through a vacuum or air, always.
For example a twisted pair in a CAT5e cable has a velocity factor of 64%, so the signal travels at 0.64c, and it will go about 8" in a nanosecond. A nanosecond is a long time in some electronic contexts.It's 4 clock cycles in a modern CPU, for example.
Any configuration of conductors of finite size has inductance and capacitance and (usually) resistance so it can be approximated using lumped components at a finer level of granularity. You might replace the wire with 20 series inductors and resistors with 20 capacitors to the ground plane. If the wavelength is very short compared to the length, you might need 200 or 2000 or .. whatever to closely approximate the wire and other methods might start to look attractive, such as transmission line theory (typically a one semester undergrad course for EEs).
"Laws" like KVL, KCL are mathematical models that approximate reality very accurately under appropriate conditions. More general laws such as Maxwell's equations apply more generally. There might be some situations (relativistic perhaps) where Maxwell's equations are no longer very accurate.
answered 1 hour ago
Spehro Pefhany
194k4139384
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Electrical signals take time to propagate through wires (and PCB traces). Slower than EM waves through a vacuum or air, always.
For example a twisted pair in a CAT5e cable has a velocity factor of 64%, so the signal travels at 0.64c, and it will go about 8" in a nanosecond. A nanosecond is a long time in some electronic contexts.It's 4 clock cycles in a modern CPU, for example.
Any configuration of conductors of finite size has inductance and capacitance and (usually) resistance so it can be approximated using lumped components at a finer level of granularity. You might replace the wire with 20 series inductors and resistors with 20 capacitors to the ground plane. If the wavelength is very short compared to the length, you might need 200 or 2000 or .. whatever to closely approximate the wire and other methods might start to look attractive, such as transmission line theory (typically a one semester undergrad course for EEs).
"Laws" like KVL, KCL are mathematical models that approximate reality very accurately under appropriate conditions. More general laws such as Maxwell's equations apply more generally. There might be some situations (relativistic perhaps) where Maxwell's equations are no longer very accurate.
answered 1 hour ago
Spehro Pefhany
194k4139384
Electrical signals take time to propagate through wires (and PCB traces). Slower than EM waves through a vacuum or air, always.
For example a twisted pair in a CAT5e cable has a velocity factor of 64%, so the signal travels at 0.64c, and it will go about 8" in a nanosecond. A nanosecond is a long time in some electronic contexts.It's 4 clock cycles in a modern CPU, for example.
Any configuration of conductors of finite size has inductance and capacitance and (usually) resistance so it can be approximated using lumped components at a finer level of granularity. You might replace the wire with 20 series inductors and resistors with 20 capacitors to the ground plane. If the wavelength is very short compared to the length, you might need 200 or 2000 or .. whatever to closely approximate the wire and other methods might start to look attractive, such as transmission line theory (typically a one semester undergrad course for EEs).
"Laws" like KVL, KCL are mathematical models that approximate reality very accurately under appropriate conditions. More general laws such as Maxwell's equations apply more generally. There might be some situations (relativistic perhaps) where Maxwell's equations are no longer very accurate.
answered 1 hour ago
Spehro Pefhany
194k4139384
answered 1 hour ago
Spehro Pefhany
194k4139384
answered 1 hour ago
Spehro Pefhany
194k4139384
answered 1 hour ago
Spehro Pefhany
194k4139384
194k4139384
add a comment |Â
add a comment |Â
up vote
1
down vote
I mean it's a direct connection, we aren't using electromagnetic waves to propogate through free space and so wavelength and stuff shouldn't matter right?
That's a very wrong assumption. The signals are still EM waves and remain EM waves, if they propagate through free space or a conductor. The laws remain the same.
At connections (wires) in the order of the length of the wavelength you can no longer used the "lumped element" approach. The "lumped element" approach means that connections are considered "ideal". For high frequency signals at distances in the order of the wavelength and larger, this approach is invalid.
So remember: the EM laws do not change as an EM wave travels through space or a conductor, they apply in both cases. EM waves remain EM waves in free space or in a conductor.
answered 1 hour ago
Bimpelrekkie
42.4k23791
Okay. I understand that the EM waves still exist when transmitting AC voltages through a wire - but they don't contribute to the actual current flow right (apart from reducing it a bit with the opposing emf). So then why should we abandon all our low-frequency and DC models when essentially the AC current is still flowing fine through that wire. I just don't see how the wavelength being too small comes into play when we have a direct wire from the AC source and load.
– AlfroJang80
1 hour ago
One should add, even for the most-high-speed signals one could expect on a "normal" PCB, the lumped model is still applicable if capacitance and inductance of a whole track are taken into account. The distances are small, after all.
– Janka
1 hour ago
@AlfroJang80: This becomes crucial as soon you are in the UHF range. Tracks acts as antennas then.
– Janka
59 mins ago
1
@AlfroJang80, a dipole antenna is just a pair of direct wires from the feed to their open ends. And yet it can transmit and receive wireless RF signals. Somewhere between a very short wire that doesn't transmit or receive any energy, and a quarter wave dipole that transmits and receives very efficiently, there must be a middle ground where radiation effects are significant but not dominant.
– The Photon
40 mins ago
Okay. Makes much more sense now. I think my confusion is that with an AC source and a wire. Is the electromagnetic field essential to current flowing? Up to now, i think of it as the emag. wave as being a seperate thing that is emitted when AC current flows through a wire. Is that correct? Or is the emag wave the only thing that makes current flow through the wire?
– AlfroJang80
2 mins ago
add a comment |Â
up vote
1
down vote
I mean it's a direct connection, we aren't using electromagnetic waves to propogate through free space and so wavelength and stuff shouldn't matter right?
That's a very wrong assumption. The signals are still EM waves and remain EM waves, if they propagate through free space or a conductor. The laws remain the same.
At connections (wires) in the order of the length of the wavelength you can no longer used the "lumped element" approach. The "lumped element" approach means that connections are considered "ideal". For high frequency signals at distances in the order of the wavelength and larger, this approach is invalid.
So remember: the EM laws do not change as an EM wave travels through space or a conductor, they apply in both cases. EM waves remain EM waves in free space or in a conductor.
answered 1 hour ago
Bimpelrekkie
42.4k23791
Okay. I understand that the EM waves still exist when transmitting AC voltages through a wire - but they don't contribute to the actual current flow right (apart from reducing it a bit with the opposing emf). So then why should we abandon all our low-frequency and DC models when essentially the AC current is still flowing fine through that wire. I just don't see how the wavelength being too small comes into play when we have a direct wire from the AC source and load.
– AlfroJang80
1 hour ago
One should add, even for the most-high-speed signals one could expect on a "normal" PCB, the lumped model is still applicable if capacitance and inductance of a whole track are taken into account. The distances are small, after all.
– Janka
1 hour ago
@AlfroJang80: This becomes crucial as soon you are in the UHF range. Tracks acts as antennas then.
– Janka
59 mins ago
1
@AlfroJang80, a dipole antenna is just a pair of direct wires from the feed to their open ends. And yet it can transmit and receive wireless RF signals. Somewhere between a very short wire that doesn't transmit or receive any energy, and a quarter wave dipole that transmits and receives very efficiently, there must be a middle ground where radiation effects are significant but not dominant.
– The Photon
40 mins ago
Okay. Makes much more sense now. I think my confusion is that with an AC source and a wire. Is the electromagnetic field essential to current flowing? Up to now, i think of it as the emag. wave as being a seperate thing that is emitted when AC current flows through a wire. Is that correct? Or is the emag wave the only thing that makes current flow through the wire?
– AlfroJang80
2 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I mean it's a direct connection, we aren't using electromagnetic waves to propogate through free space and so wavelength and stuff shouldn't matter right?
That's a very wrong assumption. The signals are still EM waves and remain EM waves, if they propagate through free space or a conductor. The laws remain the same.
At connections (wires) in the order of the length of the wavelength you can no longer used the "lumped element" approach. The "lumped element" approach means that connections are considered "ideal". For high frequency signals at distances in the order of the wavelength and larger, this approach is invalid.
So remember: the EM laws do not change as an EM wave travels through space or a conductor, they apply in both cases. EM waves remain EM waves in free space or in a conductor.
answered 1 hour ago
Bimpelrekkie
42.4k23791
I mean it's a direct connection, we aren't using electromagnetic waves to propogate through free space and so wavelength and stuff shouldn't matter right?
That's a very wrong assumption. The signals are still EM waves and remain EM waves, if they propagate through free space or a conductor. The laws remain the same.
At connections (wires) in the order of the length of the wavelength you can no longer used the "lumped element" approach. The "lumped element" approach means that connections are considered "ideal". For high frequency signals at distances in the order of the wavelength and larger, this approach is invalid.
So remember: the EM laws do not change as an EM wave travels through space or a conductor, they apply in both cases. EM waves remain EM waves in free space or in a conductor.
answered 1 hour ago
Bimpelrekkie
42.4k23791
answered 1 hour ago
Bimpelrekkie
42.4k23791
answered 1 hour ago
Bimpelrekkie
42.4k23791
answered 1 hour ago
Bimpelrekkie
42.4k23791
42.4k23791
Okay. I understand that the EM waves still exist when transmitting AC voltages through a wire - but they don't contribute to the actual current flow right (apart from reducing it a bit with the opposing emf). So then why should we abandon all our low-frequency and DC models when essentially the AC current is still flowing fine through that wire. I just don't see how the wavelength being too small comes into play when we have a direct wire from the AC source and load.
– AlfroJang80
1 hour ago
One should add, even for the most-high-speed signals one could expect on a "normal" PCB, the lumped model is still applicable if capacitance and inductance of a whole track are taken into account. The distances are small, after all.
– Janka
1 hour ago
@AlfroJang80: This becomes crucial as soon you are in the UHF range. Tracks acts as antennas then.
– Janka
59 mins ago
1
@AlfroJang80, a dipole antenna is just a pair of direct wires from the feed to their open ends. And yet it can transmit and receive wireless RF signals. Somewhere between a very short wire that doesn't transmit or receive any energy, and a quarter wave dipole that transmits and receives very efficiently, there must be a middle ground where radiation effects are significant but not dominant.
– The Photon
40 mins ago
Okay. Makes much more sense now. I think my confusion is that with an AC source and a wire. Is the electromagnetic field essential to current flowing? Up to now, i think of it as the emag. wave as being a seperate thing that is emitted when AC current flows through a wire. Is that correct? Or is the emag wave the only thing that makes current flow through the wire?
– AlfroJang80
2 mins ago
add a comment |Â
Okay. I understand that the EM waves still exist when transmitting AC voltages through a wire - but they don't contribute to the actual current flow right (apart from reducing it a bit with the opposing emf). So then why should we abandon all our low-frequency and DC models when essentially the AC current is still flowing fine through that wire. I just don't see how the wavelength being too small comes into play when we have a direct wire from the AC source and load.
– AlfroJang80
1 hour ago
One should add, even for the most-high-speed signals one could expect on a "normal" PCB, the lumped model is still applicable if capacitance and inductance of a whole track are taken into account. The distances are small, after all.
– Janka
1 hour ago
@AlfroJang80: This becomes crucial as soon you are in the UHF range. Tracks acts as antennas then.
– Janka
59 mins ago
1
@AlfroJang80, a dipole antenna is just a pair of direct wires from the feed to their open ends. And yet it can transmit and receive wireless RF signals. Somewhere between a very short wire that doesn't transmit or receive any energy, and a quarter wave dipole that transmits and receives very efficiently, there must be a middle ground where radiation effects are significant but not dominant.
– The Photon
40 mins ago
Okay. Makes much more sense now. I think my confusion is that with an AC source and a wire. Is the electromagnetic field essential to current flowing? Up to now, i think of it as the emag. wave as being a seperate thing that is emitted when AC current flows through a wire. Is that correct? Or is the emag wave the only thing that makes current flow through the wire?
– AlfroJang80
2 mins ago
Okay. I understand that the EM waves still exist when transmitting AC voltages through a wire - but they don't contribute to the actual current flow right (apart from reducing it a bit with the opposing emf). So then why should we abandon all our low-frequency and DC models when essentially the AC current is still flowing fine through that wire. I just don't see how the wavelength being too small comes into play when we have a direct wire from the AC source and load.
– AlfroJang80
1 hour ago
Okay. I understand that the EM waves still exist when transmitting AC voltages through a wire - but they don't contribute to the actual current flow right (apart from reducing it a bit with the opposing emf). So then why should we abandon all our low-frequency and DC models when essentially the AC current is still flowing fine through that wire. I just don't see how the wavelength being too small comes into play when we have a direct wire from the AC source and load.
– AlfroJang80
1 hour ago
One should add, even for the most-high-speed signals one could expect on a "normal" PCB, the lumped model is still applicable if capacitance and inductance of a whole track are taken into account. The distances are small, after all.
– Janka
1 hour ago
One should add, even for the most-high-speed signals one could expect on a "normal" PCB, the lumped model is still applicable if capacitance and inductance of a whole track are taken into account. The distances are small, after all.
– Janka
1 hour ago
@AlfroJang80: This becomes crucial as soon you are in the UHF range. Tracks acts as antennas then.
– Janka
59 mins ago
@AlfroJang80: This becomes crucial as soon you are in the UHF range. Tracks acts as antennas then.
– Janka
59 mins ago
1
1
@AlfroJang80, a dipole antenna is just a pair of direct wires from the feed to their open ends. And yet it can transmit and receive wireless RF signals. Somewhere between a very short wire that doesn't transmit or receive any energy, and a quarter wave dipole that transmits and receives very efficiently, there must be a middle ground where radiation effects are significant but not dominant.
– The Photon
40 mins ago
@AlfroJang80, a dipole antenna is just a pair of direct wires from the feed to their open ends. And yet it can transmit and receive wireless RF signals. Somewhere between a very short wire that doesn't transmit or receive any energy, and a quarter wave dipole that transmits and receives very efficiently, there must be a middle ground where radiation effects are significant but not dominant.
– The Photon
40 mins ago
Okay. Makes much more sense now. I think my confusion is that with an AC source and a wire. Is the electromagnetic field essential to current flowing? Up to now, i think of it as the emag. wave as being a seperate thing that is emitted when AC current flows through a wire. Is that correct? Or is the emag wave the only thing that makes current flow through the wire?
– AlfroJang80
2 mins ago
Okay. Makes much more sense now. I think my confusion is that with an AC source and a wire. Is the electromagnetic field essential to current flowing? Up to now, i think of it as the emag. wave as being a seperate thing that is emitted when AC current flows through a wire. Is that correct? Or is the emag wave the only thing that makes current flow through the wire?
– AlfroJang80
2 mins ago
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up vote
1
down vote
They don't break down, but when the rise time approaches 10% or is less than the propagation delay to a load impedance matching is important due to that wavelength. Load Impedance is inverted to a source at 1/4 wavelength whether it is conducted or radiated.
If the load is not matched impedance to the "transmission line and source" reflections will occur according to some coefficient called return loss and the reflection coefficient.
Here is an experiment you can do to demonstrate conducted EM waves.
If you try probing a 1MHz square wave on a 10:1 scope probe with the 10cm ground clip you might see 20MHz lumped coax resonance. Yes, the probe is not matched to the 50 Ohm generator so reflections will occur according to the 10nH/cm ground lead and 50pF/m special probe coax. Still a lumped element (LC) response.
Reducing the 10:1 probe to <1cm to just the pin tip and ring without long ground clip, raises the resonant frequency perhaps to the limitation of the probe and scope at 200MHz.
NOw try a 1:1 1m coax which is 20ns/m so a 20~50MHz square wave on a 1m coax with a 1:1 probe will see a reflection at one fraction of a wavelength and horrible square wave response unless terminated at the scope with 50 Ohms. This is a conducted EM wave reflection.
But consider a fast logic signal with 1ns rise time may have 25 Ohm source impedance and it has a >300MHz BW so overshoot can be a measurement error or actual impedance mismatch with track length reflections.
Now compute 5% of the wavelength of 300MHz at 3e8m/s for air and 2e8m/s for coax and see what the propagation delay times are that cause echoes from mismatched load e.g. CMOS high Z and say 100-ohm tracks. This is why controlled impedances are needed usually above 20~50MHz and this as an effect on ringing or overshoot or impedance mismatch. But without, this is why logic has a such a large grey zone between "0 & 1" to allow for some ringing.
If any words are unknown, look them up.
answered 1 hour ago
Tony EE rocketscientist
57.6k22084
add a comment |Â
up vote
1
down vote
They don't break down, but when the rise time approaches 10% or is less than the propagation delay to a load impedance matching is important due to that wavelength. Load Impedance is inverted to a source at 1/4 wavelength whether it is conducted or radiated.
If the load is not matched impedance to the "transmission line and source" reflections will occur according to some coefficient called return loss and the reflection coefficient.
Here is an experiment you can do to demonstrate conducted EM waves.
If you try probing a 1MHz square wave on a 10:1 scope probe with the 10cm ground clip you might see 20MHz lumped coax resonance. Yes, the probe is not matched to the 50 Ohm generator so reflections will occur according to the 10nH/cm ground lead and 50pF/m special probe coax. Still a lumped element (LC) response.
Reducing the 10:1 probe to <1cm to just the pin tip and ring without long ground clip, raises the resonant frequency perhaps to the limitation of the probe and scope at 200MHz.
NOw try a 1:1 1m coax which is 20ns/m so a 20~50MHz square wave on a 1m coax with a 1:1 probe will see a reflection at one fraction of a wavelength and horrible square wave response unless terminated at the scope with 50 Ohms. This is a conducted EM wave reflection.
But consider a fast logic signal with 1ns rise time may have 25 Ohm source impedance and it has a >300MHz BW so overshoot can be a measurement error or actual impedance mismatch with track length reflections.
Now compute 5% of the wavelength of 300MHz at 3e8m/s for air and 2e8m/s for coax and see what the propagation delay times are that cause echoes from mismatched load e.g. CMOS high Z and say 100-ohm tracks. This is why controlled impedances are needed usually above 20~50MHz and this as an effect on ringing or overshoot or impedance mismatch. But without, this is why logic has a such a large grey zone between "0 & 1" to allow for some ringing.
If any words are unknown, look them up.
answered 1 hour ago
Tony EE rocketscientist
57.6k22084
add a comment |Â
up vote
1
down vote
up vote
1
down vote
They don't break down, but when the rise time approaches 10% or is less than the propagation delay to a load impedance matching is important due to that wavelength. Load Impedance is inverted to a source at 1/4 wavelength whether it is conducted or radiated.
If the load is not matched impedance to the "transmission line and source" reflections will occur according to some coefficient called return loss and the reflection coefficient.
Here is an experiment you can do to demonstrate conducted EM waves.
If you try probing a 1MHz square wave on a 10:1 scope probe with the 10cm ground clip you might see 20MHz lumped coax resonance. Yes, the probe is not matched to the 50 Ohm generator so reflections will occur according to the 10nH/cm ground lead and 50pF/m special probe coax. Still a lumped element (LC) response.
Reducing the 10:1 probe to <1cm to just the pin tip and ring without long ground clip, raises the resonant frequency perhaps to the limitation of the probe and scope at 200MHz.
NOw try a 1:1 1m coax which is 20ns/m so a 20~50MHz square wave on a 1m coax with a 1:1 probe will see a reflection at one fraction of a wavelength and horrible square wave response unless terminated at the scope with 50 Ohms. This is a conducted EM wave reflection.
But consider a fast logic signal with 1ns rise time may have 25 Ohm source impedance and it has a >300MHz BW so overshoot can be a measurement error or actual impedance mismatch with track length reflections.
Now compute 5% of the wavelength of 300MHz at 3e8m/s for air and 2e8m/s for coax and see what the propagation delay times are that cause echoes from mismatched load e.g. CMOS high Z and say 100-ohm tracks. This is why controlled impedances are needed usually above 20~50MHz and this as an effect on ringing or overshoot or impedance mismatch. But without, this is why logic has a such a large grey zone between "0 & 1" to allow for some ringing.
If any words are unknown, look them up.
answered 1 hour ago
Tony EE rocketscientist
57.6k22084
They don't break down, but when the rise time approaches 10% or is less than the propagation delay to a load impedance matching is important due to that wavelength. Load Impedance is inverted to a source at 1/4 wavelength whether it is conducted or radiated.
If the load is not matched impedance to the "transmission line and source" reflections will occur according to some coefficient called return loss and the reflection coefficient.
Here is an experiment you can do to demonstrate conducted EM waves.
If you try probing a 1MHz square wave on a 10:1 scope probe with the 10cm ground clip you might see 20MHz lumped coax resonance. Yes, the probe is not matched to the 50 Ohm generator so reflections will occur according to the 10nH/cm ground lead and 50pF/m special probe coax. Still a lumped element (LC) response.
Reducing the 10:1 probe to <1cm to just the pin tip and ring without long ground clip, raises the resonant frequency perhaps to the limitation of the probe and scope at 200MHz.
NOw try a 1:1 1m coax which is 20ns/m so a 20~50MHz square wave on a 1m coax with a 1:1 probe will see a reflection at one fraction of a wavelength and horrible square wave response unless terminated at the scope with 50 Ohms. This is a conducted EM wave reflection.
But consider a fast logic signal with 1ns rise time may have 25 Ohm source impedance and it has a >300MHz BW so overshoot can be a measurement error or actual impedance mismatch with track length reflections.
Now compute 5% of the wavelength of 300MHz at 3e8m/s for air and 2e8m/s for coax and see what the propagation delay times are that cause echoes from mismatched load e.g. CMOS high Z and say 100-ohm tracks. This is why controlled impedances are needed usually above 20~50MHz and this as an effect on ringing or overshoot or impedance mismatch. But without, this is why logic has a such a large grey zone between "0 & 1" to allow for some ringing.
If any words are unknown, look them up.
answered 1 hour ago
Tony EE rocketscientist
57.6k22084
edited 48 mins ago
answered 1 hour ago
Tony EE rocketscientist
57.6k22084
answered 1 hour ago
Tony EE rocketscientist
57.6k22084
answered 1 hour ago
Tony EE rocketscientist
57.6k22084
57.6k22084
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